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r-ruslan [8.4K]

2 years ago

6

C is the point (-301, 601) Does C lie on the straight line passing through A and B?

1 answer:

Marat540 [252]2 years ago

8 0

Answer:

There is no picture or graph to go with the question so I am afraid I will not be able to give you a specific answer.

To find out if a point (x, y) is on the graph of a line, we plug in the values into that equation and see if we get a true statement, such as 10 = 10. If we get something different, like 6 = 4, we know that the point is not on the line because it does not satisfy the equation. Plug in (-301, 601) into the equation of the line to see whether that point lies on it or not.

Step-by-step explanation:

Suppose the equation of the straight line that passes through E and F is this:

y = 7x + 2

We are to figure out whether or not the point (1, 10) lies on that line. In order to do this we would plug in (1, 10) into the equation, with 1 being x and 10 being y.

10 = 7(1) + 2 = 7 + 2 = 9

10 = 9 is a false statement. Therefore, the point (1, 10) does NOT lie on the line y = 7x + 2.

If you were to provide an image or graph that shows the equation of line AB then perhaps I would be able to answer your question with a specific answer.

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3 years ago

Is 4x positive or negative?

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3 years ago

Thomas wants to invite Madeline to a party. He has an 80% chance of bumping into her at school. Otherwise, he'll call her on the

mash [69]

Answer:

The probability of Thomas inviting Madeline to the party over the phone is 0.143.

Step-by-step explanation:

Consider the tree diagram below.

The events are denoted as follows:

<em>A</em> = Thomas bumps into Madeline at school

<em>B</em> = Thomas call Madeline on the phone

<em>X</em> = Thomas asks Madeline to the party

The information provided is:

P (A) = 0.80

P (B) = 1 - P (A) = 1 - 0.80 = 0.20

P (X|A) = 0.90

⇒ P (X'|A) = 1 - P (X|A) = 1 - 0.90 = 0.10

P (X|B) = 0.60

⇒ P (X'|B) = 1 - P (X|B) = 1 - 0.60 = 0.40

The conditional probability of event <em>U</em> given that another events <em>V</em> has already occurred is:

$P(U|V)=\frac{P(V|U)P(U)}{P(V)}$

The law of total probability states that:

$P(V)=P(V|U)P(U)+P(V|U')P(U')$

In this case we need to determine the probability that Thomas invites Madeline to the party over the phone, i.e. P (B|X).

Use the law of total probability to determine the value of P (X) as follows:

$P(X) = P(X|A)P(A)+P(X|B)P(B)$

$=(0.90\times 0.80)+(0.60\times 0.20)\\=0.72+0.12\\=0.84$

Compute the value of P (B|X) as follows:

$P(B|X)=\frac{P(X|B)P(B)}{P(X)}$

$=\farc{0.60\times 0.20}{0.84}\\\\=0.14286\\\\\approx 0.143$

Thus, the probability of Thomas inviting Madeline to the party over the phone is 0.143.

6 0

4 years ago

What is the solution to 3 + 4e^x+1 = 11?

Alexandra [31]

What is the solution to 3 + 4e^x+1 = 11?

Solution:

$3+4e^{x+1}=11$

To solve for x, Let us subtract 3 from both sides

$3+4e^{x+1}-3=11-3$

$0+4e^{x+1}=8$

$4e^{x+1}=8$

Let us divide by 4 on both sides

$\frac{4e^{x+1}}{4} =\frac{8}{4}$

$\frac{1e^{x+1}}{1} =\frac{2}{1}$

$e^{x+1} =2$

Let us take ln on both sides

$ln (e^{x+1}) =ln(2)$

So, x+1=ln(2)

To solve for x, Let us subtract 1 from both sides

x+1-1=ln(2) -1

x+0=ln(2)-1

x= ln(2) -1

Answer: Option (A)

x=ln(2)-1

8 0

3 years ago

Read 2 more answers